Maths Problem

[grasscarp]

I can't find a special section for Maths questions in Topics, so will post it here in the hope that someone can help. It is for a child whose teacher would not explain it to him (or give the answer!)

2x over x-1 minus 7x -3 over x squared - 1

The answer and bit of guidance with how to tackle the problem would be much appreciated.

[piggynose]

I´m sure molly knows

[piggynose]

seriously though, i´m going to look in 1 of my old maths books, and join the bright sparks later

[quinie]

I got the answer one quarter (1 over 4) but I'm still working out how to explain how i did it

[mickrog]

show the equation again as unclear...are there brackets?

[Old_Geezer]

If I can think back that far, with these sorts of questions I try multiplying throughout by the denominators, in order to get rid of the fractions. Then rearranging the terms obtained to try to make a quadratic, which is solved by any one of a number of ways, but commonly by using the usual equation.

[jsnhghs]

LCM is x^2 -1

adding fractions gives 2x(x +1) -(7x-3) all over x^2-1

giving 2x^2 +2x -7x + 3

final answer (2x +1)(x-3)/(x^2-1)

Jason

[Old_Geezer]

Format will probably screw but ...

x = -b ± sqrt(b^2 - 4ac)
--------------------
2a

[grasscarp]

Mickrog there aren't any brackets. I will print off all answers to show my young friend. I never did like maths although my mother was a maths teacher. It must have skipped a generation as my daughter did pure maths at degree level!

[Ellipsis]

That can't be solved as there's no equals sign. If we assume it all equals zero, then the answers are x=-0.5 or x=3.

I'll give you the working later ...

[Old_Geezer]

:-) You are quite right. I assumed = 0.

[dr b]

assuming (2x)/(x-1) - (7x-3)/(x^2-1),

x^2 -1 = (X+1)(x-1)

so:

2x(x+1)/((x+1)(x-1) - (7x-3)/(x+1)(x-1) (mult. 1st term by (x+1)/(x+1) = 1


so:

[2x^2 + 2x - 7x + 3]/(x+1)(x-1) =

2x^2 - 5x +3//(x+1)(x-1) =

= (2x-3)(x-1)/(x+1)(x-1)

= (2x-3)/(x+1)

[Ellipsis]

Oh, the working is as jsnhghs posted (although he used quite a bit of shorthand). I thought you were trying to solve for x, but he's reduced the equation to its simplest form (without solving for x, because you can't solve for x with what you posted grasscarp).

[dr b]

I assumed the question was just asking for the expression to be simplified

[quinie]

grasscarp, I don't think very much of the child's teacher! Their job is to teach and explain things when a child asks for help

[jsnhghs]

Apologies, mistake in my factorisation.

dr b's version is correct

[quinie]

I worked it out by saying let x=2 and got 4 over 1 minus11 over 4, then got 16 over 4 minus11 over 4which =5 over 4 =1 and 1quarter. take away the-1 at the end and that leaves one quarter

[grasscarp]

I think it is a case of simplification. It is not seeking to find what x is because, as you say there is no equals in this. I really appreciate all the input. I agree with you quinie, the teacher should be helping the boy, unless he doesnt really know himself?

[piggynose]

omg!! quinie, the value of x was not asked for,
stand in the corner and be a good student

lol

[quinie]

I wasn't giving the value of x I thought this was just asking for it to be solved. I give up and leave it to the experts- no maths teachers out there?

[bibblebub]

use a common denominator of (x^2-1) = (x-1)(x+1)

2x/(x-1) = 2x(x+1)/((x-1)(x+1) = (2x^2+2x)/((x-1)(x+1))

1 = (x^2-1)/((x-1)(x+1))

thus the sum is (collecting all the numerators)

((2x^2+2x - (7x-3) - (x^2-1)) / ((x-1)(x+1))

= (2x^2 + 2x - 7x + 3 - x^2 + 1)/((x-1)(x+1))

= (x^2 - 5x + 4)/((x-1)(x+1)) =

= (x-1)(x+5)/((x-1)(x+1)) = (x+5)/(x+1)